Question

A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

Answer 1

Capacitance of the capacitor, C = 100 μF = 100 × 10⁻⁶ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

(a) Frequency of oscillations, v = 60 Hz

Angular frequency, ω = 2πv = 2π × 60 rad/s 

For an RC circuit, we have the relation for impedance as:

Z = `"R"^2 + 1/(ω^2"C"^2)`

Peak voltage, V₀ = `"V"sqrt2 = 110 sqrt2 "V"`

Maximum current is given as:

I₀ = `"V"_0/"Z"`

= `"V"_0/sqrt("R"^2 + 1/(ω^2"C"^2))`

= `(110 sqrt2)/sqrt((40)^2 + 1/((120π)^2 xx (10^-4)^2))`

= `(110 sqrt2)/sqrt(1600 + 10^8/(120π)^2)`

= 3.24 A

(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of Φ. This angle is given by the relation:

∴ `tan phi = (1/(ω"C"))/"R" = 1/(ω"CR")`

= `1/(120π xx 10^-4 xx 10)`

= 0.6635

`phi` = tan⁻¹ (0.6635) = 33.56°

= `(33.56π)/180 "rad"`

∴ Time lag = `phi/ω`

= `(33.56π)/(180 xx 120π)`

= 1.55 × 10⁻³ s

= 1.55 ms

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.